1
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 16th April Morning Slot

The relative uncertainly in the period of a satellite orbiting around the earth is 10-2. If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth is :
A
10$$-$$2
B
2 $$ \times $$ 10$$-$$2
C
3 $$ \times $$ 10$$-$$2
D
6 $$ \times $$ 10$$-$$2

Explanation

From kepler's law,

T = 2$$\pi $$ $$\sqrt {{{{r^3}} \over {GM}}} $$

$$ \Rightarrow $$$$\,\,\,$$T2 = $${{4{\pi ^2}} \over {GM}}{r^3}$$

$$ \Rightarrow $$$$\,\,\,$$ M = $${{4{\pi ^2}} \over G} \times {{{r^3}} \over {{T^2}}}$$

$$\therefore\,\,\,$$ $${{\Delta M} \over M}$$ = 2 $${{\Delta T} \over T}$$ + 3 $${{\Delta r} \over r}$$

as $${{\Delta r} \over r} \simeq 0$$

$$\therefore\,\,\,$$ $$\left| {{{\Delta M} \over M}} \right|$$ = 2 $${{\Delta T} \over T}$$ = 2 $$ \times $$ 10$$-$$2
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

A copper wire is stretched to make it 0.5% longer. The percentage change in its electrical resistance if its volume remains unchanged is :
A
2.0 %
B
2.5 %
C
1.0 %
D
0.5 %

Explanation

We know,

$$R = {{\rho l} \over A}$$

and Volume (V) = A$$l$$

$$ \Rightarrow $$   A $$=$$ $${V \over l}$$

$$ \therefore $$   $$R = {{\rho {l^2}} \over v}$$

$$ \therefore $$   $${{\Delta R} \over R} = 2{{\Delta l} \over l}$$

$$=$$   2 $$ \times $$ 0.5

$$=$$ 1%
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

Expression for time in terms of G(universal gravitional constant), h (Planck constant) and c (speed of light) is proportional to :
A
$$\sqrt {{{h{c^5}} \over G}} $$
B
$$\sqrt {{{{c^3}} \over {Gh}}} $$
C
$$\sqrt {{{Gh} \over {{c^5}}}} $$
D
$$\sqrt {{{Gh} \over {{c^3}}}} $$

Explanation

Let t $$ \propto $$ Gx hy cz

$$ \therefore $$   [t] = [G]x [h]y [c]z    . . . . . (1)

We know,

F = $${{G{M^2}} \over {{R^2}}}$$

$$ \Rightarrow $$  G = $${{F{R^2}} \over {{M^2}}}$$

$$ \therefore $$  [G] = $${{\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]} \over {\left[ {{M^2}} \right]}}$$

[G] = $$[{M^{ - 1}}{L^3}{T^{ - 2}}]$$

Also,

E = hf

$$ \therefore $$  [h] = $${{[E]} \over {[F]}}$$

= $${{\left[ {M{L^2}{T^{ - 2}}} \right]} \over {[{T^{ - 1}}]}}$$

= $$\left[ {M{L^2}{T^{ - 1}}} \right]$$

[C] = $$\left[ {{M^o}L\,{T^{ - 1}}} \right]$$

From equation (1) we get,

$$\left[ {{M^o}\,{L^o}\,{T^1}} \right] = {\left[ {{M^{ - 1}}\,{L^3}\,{T^{ - 2}}} \right]^x}{\left[ {M\,{L^2}\,{T^{ - 1}}} \right]^y}{\left[ {{M^o}L{T^{ - 1}}} \right]^z}$$

$$\left[ {{M^o}\,{L^o}\,{T^1}} \right] = \left[ {{M^{ - x + y}}\,{L^{3x + 2yz}}\,{T^{ - 2x - y - z}}} \right]$$

By comparing the power of M, L, T

$$-$$ x + y = 0

$$ \Rightarrow $$  x = y

3x + 2y + z = 0

$$ \Rightarrow $$  5x + z = 0 . . . . . (2)

$$-$$ 2x $$-$$ y $$-$$ z = 1

$$ \Rightarrow $$  $$-$$ 3x $$-$$ z = 1  . . . . (3)

By solving (2) and (3), we get,

x = $${1 \over 2}$$ = y and z = $$-$$ $${5 \over 2}$$

$$ \therefore $$  t $$ \propto $$ $$\sqrt {{{Gh} \over {{C^5}}}} $$
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line.

The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is :
A
5.755 mm
B
5.950 mm
C
5.725 mm
D
5.740 mm

Explanation

We know,

Least count (LC) = $${{Pitch} \over {no.\,of\,divisions}}$$

$$ \therefore $$  LC = $${{0.5} \over {100}}$$

= 0.5 $$ \times $$ 10$$-$$2 mm

Reading = MSR + CSR $$-$$ positive error

Given, Main scale reading (MSR) = 5.5 mm

Circular scale reading (CSR)

= 48 $$ \times $$ 0.5 $$ \times $$ 10$$-$$2 mm

= 0.24

As zero of its circular scale lines 3 division below the mean line, it means error is position error.

$$ \therefore $$  positive error

= 3 $$ \times $$ 0.5 $$ \times $$ 10$$-$$2 mm

= 0.015 mm

$$ \therefore $$  Reading = 5.5 + 0.24 $$-$$ 0.015

= 5.725 mm

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